In the estimation of nitrogen by Kjeldahl's method 0.933 g of an organic compound ' X ' was analyzed. Ammonia evolved was absorbed in 60 mL of 0.1MH2SO4. The unreacted acid requires 20 mL of 0.1 M NaOH for complete neutralization. The compound ' X ' is
The organic compound X is C6H5NH2 (aniline). Here's a concise breakdown of the solution. We're using the Kjeldahl method to find the nitrogen content of compound X. (A) Moles of acid reacted with ammonia Total H2SO4 added ; 0.060L×0.1M=0.006mol. NaOH used in back-titration : 0.020I×0.1M=0.002mol. Unreacted H2SO4 (from 2NaOH:1H2SO4 ) : 0.002mol∕2=0.001mol. H2SO4 reacted with NH3:0.006mol0.001mol=0.005mol. (B) Mass of nitrogen in compound X Moles of NH3 (from 2NH3 : 1H2SO4 ) : 0.005mol×2=0.010mol. Mass of nitrogen ( N=14.01g∕mol ) : 0.010mol×14.01g∕mol=0.1401g. (C) Percentage of nitrogen in compound X. Sample mass =0.933g %N=(0.1401g∕0.933g)×100% ≈15.01%. (D) Compare with options Aniline (C6H5NH2) : Contains one N . Molar mass ≈93g∕mol.%N=(14∕93)×100%≈15.05%. Other options (benzylamine, n-propylamine, acetamide) have significantly different nitrogen percentages (approx. 13.08\%,23.73%,23.73% respectively). Since compound X has approximately 15.01% nitrogen, it matches Aniline.